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I'm following Griffiths' intro QM text, 2nd edition. We have defined the angular momentum operator and obtained the commutation relations $$[L_i, L_j] = i\hbar\epsilon_{ijk} L_k$$. We notice in particular how the components are incompatible observables so there's no sense in trying to simultaneously diagonalising them. Instead, after noticing that $$[L_i, L^2] = 0$$, we try to simultaneously diagonalise $$L^2$$ and one component, say $$L_z$$.

We define the ladder operators $$L_\pm = L_x + iL_y$$, and observe that, since $$[L^2, L_\pm] = 0$$ and $$[L_z, L_\pm] = \pm\hbar L_\pm$$, if $$\psi$$ is a simultaneous eigenstate of $$L^2, L_z$$ with corresponding eigenvalues $$\lambda,\mu$$, then $$L_\pm\psi$$ is also a simultaneous eigenstate of $$L^2, L_z$$ with corresponding eigenvalues $$\lambda, \mu\pm\hbar$$.

Thus if one simultaneous eigenstate $$\psi$$ is known, from it we can obtain a whole sequence of simultaneous eigenstates by repeated application of the ladder operators. We observe however that for any given eigenstate $$\lambda > \mu^2$$, so that this method of construction of new states must fail at some point. We conclude that there must exist a "top" state $$\psi_+$$ (or "bottom" state $$\psi_-$$) such that $$L_\pm\psi_\pm$$, the result of raising/lowering it, cannot be normalisable.

At this point, Griffiths starts an argument to relate the eigenvalues $$\lambda$$ and $$\mu$$ (the conclusion is that $$\mu_\pm = \pm\hbar l$$ for some positive integer or half-integer $$l$$, and $$\lambda=\hbar^2 l(l+1)$$). The first argument presented relies on the assumption that $$L_\pm \psi_\pm = 0$$, which as we remarked earlier is not necessary—$$L_\pm\psi_\pm$$ need only be non-normalisable.

QUESTION: How to obtain the result without this unjustified assumption $$L_\pm\psi_\pm = 0$$? i.e. either justify it or assume only non-normalisability.

A footnote in the page mentions the fact that the assumption $$L_\pm \psi_\pm = 0$$ is not completely due, and refers the reader to Problem 4.18, which is said to explore this. Using

$$L_\pm L_\mp = L^2 - L_z^2 \pm \hbar L_z \qquad\qquad L_\pm^\dagger = L_\mp$$

as suggested, I can obtain

$$|L_\pm\psi|^2 = \ = \lambda - \mu(\mu\pm\hbar)$$

but at this point Griffiths seems to assume $$\lambda=\hbar^2 l(l+1)$$ where $$\mu_\pm=\pm\hbar l$$ (which can then be used to conclude that $$|L_\pm\psi_\pm|=0$$ indeed), but that is cheating since these had been previously derived under the assumption we're trying to avoid.

It is an amazing coincidence that I was revisiting that point in Griffths book today. The reason for it was replacing $$L$$ by $$J = L + S$$. As $$J$$ follows the same commutting relations than $$L$$, I thought that the conclusions should be similar, explaining also the half integers of the solution. I am still struggling with the maths, but here is what I got to this point:

Supposing $$f$$ eigenfunction of $$J^2$$ and $$J_Z$$ and normalized, and let's examine the conditions for $$J_+f$$ also be normalized: $$ = = = = = 1$$

$$J^2f = \lambda f$$ and $$J_z^2f = J_zJ_zf = \mu^2f$$ so: $$\lambda – \mu^2 – \mu = 1$$; $$\mu^2 + \mu – \lambda + 1 = 0$$

$$\mu = -1/2 +/-(1/4 – (1 – \lambda))^{1/2} = -1/2 +/- (\lambda - 3/4)^{1/2}$$

Now, the minimum value for \lambda is $$3/4$$, because the eigenvalues are real. So, $$\mu = -1/2$$. That means: for that lambda, only a $$f$$ with $$\mu = -1/2$$ can be raised, (if the functions are normalized).

The same can be done for $$J_-f$$, and $$f$$ must have $$\mu = 1/2$$ to be lowered.

So, for $$\lambda = 3/4$$, $$\mu = -1/2$$ or $$1/2$$.

If we want to lower $$\mu$$ to $$-3/2$$, the consequence from the conditions of normalization (the equation relating $$\lambda$$ and $$\mu$$) is that $$\lambda = 7/4$$.

And if we use $$\lambda = 7/4$$ for $$J_-f$$, $$f$$ must have $$\mu = 3/2$$ to be lowered.

The next $$\lambda = 19/4$$

That expression $$(\lambda - 3/4)^{1/2}$$ assumes the values of $$l$$ $$(0, 1, 2 ...)$$.

$$3/4$$ is certainly the eigenvalue for $$S^2 = S(S+1) = (1/2)(1/2+1)$$

So,$$(\lambda - 3/4)^{1/2}$$ = $$(J^2 - S^2)^{1/2} = L$$

But I can not find the eigenvalue for $$L^2 = l(l+1)$$. Neither a general formula for the eingenvalues of $$J^2 = \lambda$$.

Edit from Abr, 27th:

The procedure above led me to nowhere. But after trying to understand to origin of $$L^2 = l(l+1)$$, only through the ladder procedure, i realized that it is impossible.

That eigenvalue results from the solution of the angular part of the Schrodinger equation for a spherical symmetric potential. That differential equation happens to be the same as the generated by:

$$L^2f = (L^+L^- + L_z^2 - L_z)f$$

The spherical harmonics that solve the resultant differential equation requires that:

$$(L^+L^- + L_z^2 - L_z)f = l(l+1)f$$

where $$l$$ is a non negative integer.